(a) Rewrite the expression for the total free energy change for nucleation (Equation 10.1) for the case of a cubic nucleus of edge length \(a\) (instead of a sphere of radius \(r\) ). Now differentiate this expression with respect to \(a\) (per Equation 10.2 ) and solve for both the critical cube edge length, \(a^{*},\) and also \(\Delta G^{*}\) (b) Is \(\Delta G^{*}\) greater for a cube or a sphere? Why?

The expression for the total free energy change of a spherical nucleus is given by Equation 10.1: \(\Delta G = - \frac{4}{3} \pi r^3 \Delta g_v + 4 \pi r^2 \gamma\) Here, \(\Delta g_v\) is the volume free energy change, \(r\) is the radius of the sphere, and \(\gamma\) is the interface energy.

To rewrite the expression for a cubic nucleus, we can replace the volume and surface areas of a sphere with the respective values for a cube. The volume of a cube is \(a^3\) and the surface area is \(6a^2\). Therefore, the expression for total free energy change of a cubic nucleus is: \(\Delta G = - a^3 \Delta g_v + 6a^2 \gamma\)

To find the critical edge length \(a^*\), we have to differentiate the expression for \(\Delta G\) with respect to \(a\) and set it equal to zero. \(\frac{d \Delta G}{da} = - 3a^2 \Delta g_v + 12a \gamma\) Now set \(\frac{d \Delta G}{da} = 0\) and solve for \(a^*\). \(0 = - 3(a^*)^2 \Delta g_v + 12a^* \gamma\) Solve for \(a^*\): \(a^* = \frac{4\gamma}{\Delta g_v}\)

Using the critical edge length \(a^*\), we can now solve for \(\Delta G^*\): \(\Delta G^* = - (a^*)^3 \Delta g_v + 6(a^*)^2 \gamma\) Plug in the value of \(a^*\): \(\Delta G^* = - (\frac{4\gamma}{\Delta g_v})^3 \Delta g_v + 6(\frac{4\gamma}{\Delta g_v})^2 \gamma\) Simplify for \(\Delta G^*\): \(\Delta G^* = \frac{32 \gamma^3}{\Delta g_v^2}\)

For a spherical nucleus, we have \(\Delta G^* = \frac{16 \pi \gamma^3}{3(\Delta g_v)^2}\). Now compare the values of \(\Delta G^*\) for a cube and a sphere by dividing their expressions: \(\frac{\Delta G^*_{cube}}{\Delta G^*_{sphere}} = \frac{\frac{32 \gamma^3}{\Delta g_v^2}}{\frac{16 \pi \gamma^3}{3(\Delta g_v)^2}}\) After simplifying the fraction, we obtain: \(\frac{\Delta G^*_{cube}}{\Delta G^*_{sphere}} = \frac{2}{\pi} \approx 0.6366\) Since the ratio is less than \(1\), the total free energy change is smaller for a cubic nucleus as compared to a spherical nucleus. The reason \(\Delta G^*\) is smaller for a cubic nucleus is that cubes have a lower surface area to volume ratio, which lowers their surface energy contribution to \(\Delta G^*\).